Exponential functions 35
Hence α=
1
θ
ln
R
R 0
=
1
1500
ln
(
6 × 103
5 × 103
)
=
1
1500
( 0. 1823215 ...)
= 1. 215477 ···× 10 −^4
Hence α=1.215× 10 −^4 ,
correct to 4 significant figures.
From above, ln
R
R 0
=αθ
hence θ=
1
α
ln
R
R 0
When R= 5. 4 × 103 ,α= 1. 215477 ...× 10 −^4 and
R 0 = 5 × 103
θ=
1
1. 215477 ...× 10 −^4
ln
(
5. 4 × 103
5 × 103
)
=
104
1. 215477 ...
( 7. 696104 ...× 10 −^2 )
= 633 ◦C,correct to the nearest degree.
Problem 18. In an experiment involving
Newton’s law of cooling, the temperatureθ(◦C) is
given byθ=θ 0 e−kt. Find the value of constantk
whenθ 0 = 56. 6 ◦C,θ= 16. 5 ◦Candt= 83 .0seconds.
Transposing θ=θ 0 e−ktgives
θ
θ 0
=e−kt
from which
θ 0
θ
=
1
e−kt
=ekt
Taking Napierian logarithms of both sides gives:
ln
θ 0
θ
=kt
from which,
k=
1
t
ln
θ 0
θ
=
1
83. 0
ln
(
56. 6
16. 5
)
=
1
83. 0
( 1. 2326486 ...)
Hencek=1.485× 10 −^2
Problem 19. The currentiamperes flowing in a
capacitor at timetseconds is given by
i= 8. 0 ( 1 −e
−t
CR), where the circuit resistanceRis
25 × 103 ohms and capacitanceCis
16 × 10 −^6 farads. Determine (a) the currentiafter
0.5seconds and (b) the time, to the nearest
millisecond, for the current to reach 6.0A. Sketch
the graph of current against time.
(a) Currenti= 8. 0 ( 1 −e
−t
CR)
= 8 .0[1−e
− 0. 5
( 16 × 10 −^6 )( 25 × 103 )]= 8. 0 ( 1 −e−^1.^25 )
= 8. 0 ( 1 − 0. 2865047 ...)= 8. 0 ( 0. 7134952 ...)
=5.71amperes
(b) Transposingi= 8. 0 ( 1 −e
−t
CR)
gives
i
8. 0
= 1 −e
−t
CR
from which, e
−t
CR= 1 −
i
8. 0
=
8. 0 −i
8. 0
Taking the reciprocal of both sides gives:
e
t
CR=^8.^0
8. 0 −i
Taking Napierian logarithms of both sides gives:
t
CR
=ln
(
8. 0
8. 0 −i
)
Hence
t=CRln
(
8. 0
8. 0 −i
)
=( 16 × 10 −^6 )( 25 × 103 )ln
(
8. 0
8. 0 − 6. 0
)
wheni= 6 .0 amperes,
i.e. t=
400
103
ln
(
8. 0
2. 0
)
= 0 .4ln4. 0
= 0. 4 ( 1. 3862943 ...)= 0 .5545s
=555ms, to the nearest millisecond.
Agraphofcurrent against timeisshowninFig.4.6.